Question

A solid body of weight 9.0 N is suspended by a string in the water. The tension in the string is 7.5 N. Find

1. the loss of weight of the solid in water,
2. the buoyant force exerted by water on the solid,
3. the volume of the body
4. the tension in the string if only half the volume of the body is immersed in water. Take the density of water as 1.0 x 10³ kg m^-3

Answer 1

Step 1: Given Data

Weight of the body in air,$W$ $=9N$

Tension in the string,$T$$=7.5N$

density of water,$p$$=1000Kg/m^3$

acceleration due to gravity,$g=10m/s^2$

Buoyant force,$F_B$

volume of the body,$V$

Step 2: Relation between the buoyant force and density

$F_B=pVg$

Step 3: Finding the value of the buoyant force

PART 1: Weight of the body $=9N$

Tension in the string= weight in water$=7.5N$

Loss of weight of solids in water $=$$weightofsolidintheair-weightofsolidinwater$

$$\begin{gathered} =9-7.5 \\ =1.5N \end{gathered}$$

PART 2: Buoyant force exerted by water on solid

$F_B$$=lossinweightofsolidinwater$

$F_B$$=1.5N$

PART 3: Volume of the body

$$\begin{gathered} F_B=pVg \\ 1.5=1000\times V\times 10 \\ V=\frac{1.5}{1000\times 10} \\ =1.5\times {10}^{-4}{m}^3 \end{gathered}$$

volume of the body = volume of the liquid displaced$=1.5\times {10}^{-4}{m}^3$

PART 4: the tension in the string if only half the volume of the body is immersed in water

upthrust on the body if half of the volume is submerged in water, $$\begin{gathered} F_B=\frac{1.5\times {10}^{-4}\times 1000\times 10}{2} \\ =0.75N \end{gathered}$$

Tension in the string,

$$\begin{gathered} T=9-0.75 \\ =8.25N \end{gathered}$$